Explanation:
The given data is as follows.
mass of water = 35.00 kg = 35 × 1000 g = 35000 g
specific heat of water = 4.186
change in temperature = 2.113
Formula to calculate heat change is as follows.
q =
Putting the given values into the above formula as follows.
q =
q =
= 309131.9 J
or, =
= 309.1 kJ
As 7.00 g of compound 'X' is giving 309.131 kJ of heat. Therefore, 1 mole of (molar mass = 70 g) will give the heat as follows.
= 3091 kJ/mol
So, the reaction equation will be as follows.
Value of = 3091 kJ/mol for the reaction.
Standard enthalpy values for formation of included atoms or molecules into this reaction are as follows.
= -393.5 kJ/mol, = 0 kJ/mol
= -241.8 kJ/mol
Formula to calculate standard heat of formation of compound is as follows.
=
3091 kJ/mol =
= -3091 kJ/mol + 1967.5 kJ/mol + 1209 kJ/mol
= 85.5 kJ/mol
or, = -85.5 kJ/mol
Thus, we can conclude that the standard heat of formation of given compound X is -85.5 kJ/mol.
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